Heron's Formula

A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?

Answer

We know that,

Each side of the equilateral triangle is equal.

Given,

Length of each side of an equilateral triangle = a cm.

Perimeter of traffic signal board (equilateral triangle) = sum of all the sides = a + a + a = 3a cm.

By formula,

Semi perimeter (s) = $\dfrac{\text{Perimeter of triangle}}{2} = \dfrac{3a}{2}$ cm.

By Heron's formula,

Area of triangle (A) = $\sqrt{s(s - a)(s - b)(s - c)}$ sq.units, where a, b and c are sides of triangle.

Substituting values we get :

$A = \sqrt{\dfrac{3a}{2} \times \Big(\dfrac{3a}{2} - a\Big) \times \Big(\dfrac{3a}{2} - a\Big) \times \Big(\dfrac{3a}{2} - a\Big)} \\[1em] = \sqrt{\dfrac{3a}{2} \times \Big(\dfrac{3a - 2a}{2}\Big) \times \Big(\dfrac{3a - 2a}{2}\Big) \times \Big(\dfrac{3a - 2a}{2}\Big)} \\[1em] = \sqrt{\dfrac{3a}{2} \times \dfrac{a}{2} \times \dfrac{a}{2} \times \dfrac{a}{2}} \\[1em] = \sqrt{\dfrac{3a^4}{16}} \\[1em] = \dfrac{\sqrt{3}}{4}a^2.$

Given,

Perimeter = 180 cm

∴ 3a = 180

⇒ a = $\dfrac{180}{3}$ = 60 cm.

Substituting value of a, we get :

$\text{Area of triangle (A)} = \dfrac{\sqrt{3}}{4}a^2 \\[1em] = \dfrac{\sqrt{3}}{4} \times 60^2 \\[1em] = \dfrac{\sqrt{3}}{4} \times 3600 \\[1em] = 900\sqrt{3}\text{ cm}^2.$

Hence, the area of the signal board is $900\sqrt{3}$ cm².

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m. The advertisements yield an earning of ₹ 5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?

Answer

Here a, b and c are the sides of the triangle.

Let a = 122 m, b = 22 m and c = 120 m

By formula,

Semi Perimeter (s) = $\dfrac{\text{Perimeter of triangle}}{2}$

s = $\dfrac{a + b + c}{2} = \dfrac{122 + 22 + 120}{2} = \dfrac{264}{2}$ = 132 m.

By Heron's formula,

Area of triangle (A) = $\sqrt{s(s - a)(s - b)(s - c)}$ sq.units

Substituting values we get :

$\text{Area of one wall} = \sqrt{132(132 - 122)(132 - 22)(132 - 120)} \\[1em] = \sqrt{132 \times 10 \times 110 \times 12} \\[1em] = \sqrt{1742400} \\[1em] = 1320 \text{ m}^2.$

We know that,

The rent of advertising per year = ₹ 5000 per m²

So,

The rent of one complete triangular wall for 1 month

= $\dfrac{\text{Rent per sq. unit} \times \text{ Area}}{12}$

= $\dfrac{(5000 \times 1320)}{12} = 11 \times 5000$ = ₹ 5,50,000.

∴ The rent of one wall for 3 months = ₹ 5,50,000 x 3 = ₹ 16,50,000.

Hence, the rent paid by the company = ₹ 16,50,000.

There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN”. If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Answer

Let a, b and c be the sides of the triangle.

Let a = 11 m, b = 6 m and c = 15 m.

By formula,

Semi Perimeter (s) = $\dfrac{\text{Perimeter of triangle}}{2} = \dfrac{a + b + c}{2}$

= $\dfrac{11 + 6 + 15}{2} = \dfrac{32}{2}$ = 16 m.

By Heron's formula,

Area of triangle (A) = $\sqrt{s(s - a)(s - b)(s - c)}$ sq.units

Substituting values we get :

$A = \sqrt{16(16 - 11)(16 - 6)(16 - 15)} \\[1em] = \sqrt{16 \times 5 \times 10 \times 1} \\[1em] = \sqrt{800} \\[1em] = 20\sqrt{2} \text{ m}^2.$

Hence, area of the wall painted in colour = $20\sqrt{2}$ m².

Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

Answer

Let a, b and c be the sides of the triangle.

Let a = 18 cm, b = 10 cm.

Given,

Perimeter = 42 cm

∴ a + b + c = 42

⇒ 18 + 10 + c = 42

⇒ 28 + c = 42

⇒ c = 42 - 28 = 14 cm.

By formula,

Semi Perimeter (s) = $\dfrac{\text{Perimeter of triangle}}{2} = \dfrac{42}{2}$ = 21 cm.

By Heron's formula,

Area of triangle (A) = $\sqrt{s(s - a)(s - b)(s - c)}$ sq.units

Substituting values we get :

$A = \sqrt{21(21 - 18)(21 - 10)(21 - 14)} \\[1em] = \sqrt{21 \times 3 \times 11 \times 7} \\[1em] = \sqrt{4851} \\[1em] = 21\sqrt{11} \text{ cm}^2$

Hence, area of triangle = $21\sqrt{11}$ cm².

Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.

Answer

Given,

Sides of a triangle are in the ratio of 12 : 17 : 25.

Let sides of the triangle are :

a = 12x, b = 17x and c = 25x.

Given,

Perimeter = 540 cm

∴ 12x + 17x + 25x = 540 cm

⇒ 54x = 540 cm

⇒ x = $\dfrac{540}{54}$

⇒ x = 10 cm

⇒ a = 12 × 10 = 120 cm,

⇒ b = 17 × 10 = 170 cm,

⇒ c = 25 × 10 = 250 cm.

Semi perimeter (s) = $\dfrac{\text{Perimeter of triangle}}{2} = \dfrac{540}{2}$ = 270 cm.

By Heron's formula,

Area of triangle (A) = $\sqrt{s(s - a)(s - b)(s - c)}$ sq.units

Substituting values we get :

$A = \sqrt{270(270 - 120)(270 - 170)(270 - 250)} \\[1em] = \sqrt{270 \times 150 \times 100 \times 20} \\[1em] = \sqrt{81000000} \\[1em] = 9000 \text{ cm}^2.$

Hence, area of triangle = 9000 cm².

An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

Answer

Length of equal sides (a and b) = 12 cm

Let third side of triangle (c) = x cm

Given,

Perimeter of triangle = 30

∴ 12 + 12 + x = 30

⇒ 24 + x = 30

⇒ x = 30 - 24

⇒ x = 6 cm.

∴ c = 6 cm.

By formula,

Semi perimeter (s) = $\dfrac{\text{Perimeter of triangle}}{2} = \dfrac{30}{2}$ = 15 cm.

By Heron's formula,

Area of triangle (A) = $\sqrt{s(s - a)(s - b)(s - c)}$ sq.units

Substituting values we get :

$A = \sqrt{15(15 - 12)(15 - 12)(15 - 6)} \\[1em] = \sqrt{15 \times 3 \times 3 \times 9} \\[1em] = \sqrt{1215} \\[1em] = 9\sqrt{15} \text{ cm}^2.$

Hence, area of triangle = $9\sqrt{15}$ cm².